Question

A 150ml sample of hydrochloric acid completely reacted with 60.0mL of a 0.100 M NaOH solution. What was the original concentration of the HCl solution

Answer 1

0.040 M is the correct answer you are looking for

Answer 2

HCl + NaOH = NaCl + H₂O

n(HCl)=c(HCl)v(HCl)
n(NaOH)=c(NaOH)v(NaOH)

n(HCl)=n(NaOH)

c(HCl)v(HCl)=c(NaOH)v(NaOH)

c(HCl)=c(NaOH)v(NaOH)/v(HCl)

c(HCl)=0.100mmol/mL*60.0mL/150mL=0.040 mmol/mL = 0.040 mol/L

0.040 M HCl